The distance from the line [tex]y=\frac{1}{6} x+6[/tex] to the given point (-6,5) is o units .
Distance Formula : the distance between the line ax+by+c=0 and the point (p,q) is given by the formula
[tex]d=\frac{|ap+bq+c|}{\sqrt{a^{2} +b^{2} } }[/tex]
In the given question the equation of line is [tex]y=\frac{1}{6} x+6[/tex]
taking LCM and solving further
[tex]y=\frac{x+36}{6} \\ \\ 6y=x+36\\ \\ x-6y+36=0[/tex] ⇒equation of the line .
Given point (-6,5)
from above values we found that a=1 , b = -6 , c=36, p=-6, q=5.
Substituting the values in distance formula we get
[tex]d=\frac{|1*(-6)-6*5+36}{\sqrt{1^{2} +(-6)^{2} } }[/tex]
[tex]=\frac{|-6-30+36|}{\sqrt{37} }[/tex]
[tex]=\frac{-36+36}{\sqrt{37} }[/tex]
[tex]=0[/tex]
Therefore , the distance from the line [tex]y=\frac{1}{6} x+6[/tex] to the given point (-6,5) is 0 units.
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